One linear static path, end to end — SWBPIPE verification

Part VI · VerificationVI–B9

Part VI

Verification

B. Elements and loads · One linear static path, end to end

A straight member four metres long is anchored at node 0. At node 1 it carries an axial force of 12.0 N, rests on a spring of 40.0 N/m in Y, and is given an imposed displacement of −0.01 m in Z. A uniform load of −2.0 N/m acts in Y along its length. The case follows one path through the solver: load preparation, lumping, supports, solution and recovery of end forces.

It then feeds in two faulty inputs and expects each to be reported as a blocking error.

Fig. VI–B9.—One linear static path, end to end

1.Inputs.

Illustrative values, taken from no standard and chosen so the arithmetic can be followed by hand. Most do not describe a real pipe; read them in any consistent set of units.

Quantity	Symbol	Value
Length	L	4.0 m
Elastic modulus	E	1500.0 Pa
Shear modulus	G	600.0 Pa
Area	A	2.0 m²
Second moment about y	I_y	1.8 m⁴
Second moment about z	I_z	2.2 m⁴
Torsion constant	J	0.9 m⁴
Node 1 axial force, global X	P_x	12.0 N
Uniform load, global Y	w	−2.0 N/m
Node 0 anchor	—	6 restrained freedoms
Node 1 spring in Y	k	40.0 N/m
Node 1 imposed displacement in Z	u_z	−0.01 m

2.Method.

The uniform load is lumped half to each end. The axial response is a single spring of stiffness EA/L, so the end displacement is the force over that stiffness, and the recovered axial end force returns the applied force.

In Y the member is a cantilever whose tip may rotate freely, so its tip stiffness is 3EI_z/L³ = 154.6875 N/m. The spring acts in parallel. The node 1 displacement is the lumped load over the combined stiffness, and the end shear is the member’s share of the load. The imposed displacement in Z and the axial force act in planes that do not couple to Y.

u_y,j = −4.0 / (154.6875 + 40.0) = −0.020545746388443017 m(1)

V_y,j = 154.6875 × u_y,j = −3.178170144462279 N(2)

3.Results.

Expected values and diagnostics
Quantity	Expected
Lumped force at node 0, Y	−4.0 N
Lumped force at node 1, Y	−4.0 N
Node 1 axial displacement	0.016 m
Node 1 displacement in Y	−0.02054575 m
Node 1 displacement in Z	−0.01 m (imposed)
Recovered local axial force at node j	12.0 N
Recovered local shear in Y at node j	−3.17817 N
Spring with no stiffness	Blocking restraint error
Load with no target	Blocking model-topology error

The tests check that:

All solved and recovered quantities are finite.
The lumped end forces match the hand values.
The axial displacement and the recovered axial force match the hand values.
The node 1 displacement in Z equals the imposed value.
The node 1 displacement in Y and the recovered shear in Y match the hand values.
A spring with no stiffness is reported as a blocking restraint error naming that spring.
A load with no target is reported as a blocking model-topology error naming that load.

What it shows. Compared with an independent hand calculation.

Path exercised. The benchmark calls the solver’s components directly: elements, loads, frame solver and stress recovery. It does not go through the program’s own model-to-solve path.

Agreement. Each computed value must match the reference within an absolute difference of 1.0 × 10⁻⁹ in the case’s own units; counts and structural outcomes must match exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.

4.Run it yourself.

cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/mechanics/Cargo.toml tp_phys_002_integrated_fixture_solves_recovers_and_maps_diagnostics

Hand calculation: validation/hand_calcs/mechanics/tp_phys_002_linear_static_integration.md. Test record, with the recorded run of 2026-07-10, with the strengthened check first run on 2026-09-22: mech-tp-phys-002-linear-static-integration.md.